Question: Kaliska is jumping rope. The vertical height of the center of her rope off the ground $R(t)$ (in $\text{cm}$ ) as a function of time $t$ (in seconds) can be modeled by a sinusoidal expression of the form $a\cdot\cos(b\cdot t)+d$. At $t=0$, when she starts jumping, her rope is $0\text{ cm}$ off the ground, which is the minimum. After $\dfrac{\pi}{12}$ seconds, it reaches a height of $60\text{ cm}$ from the ground, which is half of its maximum height. Find $R(t)$. $\textit{t}$ should be in radians. $R(t) = $
Solution: The strategy First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph. Then, we should use the given information to find the amplitude, midline, and period of the function's graph. Finally, we should find $a$, $b$, and $d$ in the expression $a\cos(b\cdot t)+d$ by considering the features we found. Converting the given information into mathematical terms At $t=0$, the bottom of Kaliska's rope is $0\text{ cm}$ from off the ground. This means the graph of the function passes through $(0,0)$. We are given that this is minimum height, which corresponds to a minimum of the graph. $\dfrac{\pi}{12}$ seconds later (which means $t=\dfrac{\pi}{12}$ ) its distance is $60\text{ cm}$ from the ground. This corresponds to the point $\left(\dfrac{\pi}{12},60\right)$. We are given that this is the half the maximum height, which corresponds to the midline of the graph. In conclusion, the graph has a minimum point at $(0,0)$ and then intersects its midline at $(\dfrac{\pi}{12},60)$. Determining the amplitude, midline, and period The midline intersection is at $y={60}$, so this is the midline. The minimum point is $60$ units below the midline, so the amplitude is ${60}$. The minimum point is $\dfrac{\pi}{12}$ units to the left of the midline intersection, so the period is $4\cdot \dfrac{\pi}{12}={\dfrac{\pi}{3}}$. [Why did we multiply by 4?] Determining the parameters in $a\cos(b\cdot t)+d$ Since the minimum at $t=0$ is followed by a midline intersection, we know that $a<0$. [How do we know that?] The amplitude is ${60}$, so $|a|={60}$. Since $a<0$, we can conclude that $a=-60$. The midline is $y={60}$, so $d=60$. The period is ${\dfrac{\pi}{3}}$, so $b=\dfrac{2\pi}{\left({\dfrac{\pi}{3}}\right)}=6$. The answer $R(t)=-60\cos\left(6t\right)+60$